Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ over positive integers. 
Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ where $x, y, z \in
 \mathbb{N}$

I've found some solutions, like $(0, 0, 1), (1, 1, 3)$ and, more general, $x=k,y=k,z=2k+1$. No idea how to prove or disprove there is no other solutions.
 A: Write this as 
$$3x + y + 1 = z^2 - (x+y)^2 = (z-x-y)(z+x+y)$$
For any integers $a,b$, we might look for a solution with
$$ \eqalign{a &= z - x - y\cr
            b &= z + x + y\cr
           ab &= 3x + y + 1\cr} $$
Solving this system for $x,y,z$:
$$ \eqalign{x &= \frac{ab}{2} + \frac{a}{4} - \frac{b}{4} - \frac{1}{2}\cr
            y &= \frac{-ab}{2} - \frac{3a}{4} + \frac{3b}{4} + \frac{1}{2}\cr
            z &= \frac{b}{2} + \frac{a}{2}\cr}$$
In order for $x, y, z \ge 0$, we need $b > 0$ with 
$$ \frac{b+2}{2b+1} \le a  \le \frac{3b+2}{2b+3} $$
The only integer that fits is $a = 1$.  Then for $x$ to be an integer we need $b \equiv 1 \mod 4$.  The result is that all the positive integer solutions are the ones you found.
A: Actually there is a very simple solution, using the inequalities:
$(x+y)^2 \lt (x+y)^2 + 3x + y + 1 \lt (x+y+2)^2$
It follows that $(x+y)^2 + 3x + y + 1=(x+y+1)^2$ and from here $x=y$. 
A: If we set $x+y=t$, we are just looking for the solutions of 
$$ (2t+1)^2+(8x+3) = 4z^2 \tag{1}$$
or:
$$ (8x+3) = (2z+2t+1)(2z-2t-1).\tag{2} $$
If $d$ is a divisor of $8x+3$ that is $\leq \sqrt{8x+3}$, $d$ is odd and
$$ 2z+2t+1=\frac{8x+3}{d},\qquad 2z-2t-1 = d \tag{3} $$
leads to the solution:
$$ z = \frac{1}{4}\left(d+\frac{8x+3}{d}\right),\qquad 2t+1=\frac{1}{2}\left(\frac{8x+3}{d}-d\right).\tag{4} $$
I leave to you to fill the missing details, like, for instance, to find the conditions that ensure the last $z$ and $t$ actually being positive integers with $t\geq x$.
